feat: Add some basic (but probably uncomplete) schemes

This commit is contained in:
Samuel Ortion 2023-09-22 23:48:46 +02:00
parent a92a13d354
commit ae51323390
12 changed files with 191 additions and 85 deletions

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@ -13,7 +13,7 @@ with $g$ being
\subsection{Penalized Regression} \subsection{Penalized Regression}
When the number of variables is large, e.g, when the number of explicative variable is above the number of observations, if $p >> n$ ($p$: the number of explicative variable, $n$ is the number of observations), we cannot estimate the parameters. When the number of variables is large, e.g, when the number of explanatory variable is above the number of observations, if $p >> n$ ($p$: the number of explanatory variable, $n$ is the number of observations), we cannot estimate the parameters.
In order to estimate the parameters, we can use penalties (additional terms). In order to estimate the parameters, we can use penalties (additional terms).
Lasso regression, Elastic Net, etc. Lasso regression, Elastic Net, etc.
@ -21,8 +21,7 @@ Lasso regression, Elastic Net, etc.
\subsection{Simple Linear Model} \subsection{Simple Linear Model}
\begin{align*} \begin{align*}
\Y &= \X & \beta & + & \varepsilon.\\ \Y &= \X \beta + \varepsilon \\
n \times 1 & n \times 2 & 2 \times 1 & + & n \times 1 \\
\begin{pmatrix} \begin{pmatrix}
Y_1 \\ Y_1 \\
Y_2 \\ Y_2 \\
@ -35,11 +34,11 @@ Lasso regression, Elastic Net, etc.
\vdots & \vdots \\ \vdots & \vdots \\
1 & X_n 1 & X_n
\end{pmatrix} \end{pmatrix}
& \begin{pmatrix} \begin{pmatrix}
\beta_0 \\ \beta_0 \\
\beta_1 \beta_1
\end{pmatrix} \end{pmatrix}
& + & +
\begin{pmatrix} \begin{pmatrix}
\varepsilon_1 \\ \varepsilon_1 \\
\varepsilon_2 \\ \varepsilon_2 \\
@ -61,8 +60,35 @@ Lasso regression, Elastic Net, etc.
\item Graphical representation; \item Graphical representation;
\item ... \item ...
\end{enumerate} \end{enumerate}
\[
Y = X \beta + \varepsilon,
\]
is noted equivalently as
\[
\begin{pmatrix}
y_1 \\
y_2 \\
y_3 \\
y_4
\end{pmatrix}
= \begin{pmatrix}
1 & x_{11} & x_{12} \\
1 & x_{21} & x_{22} \\
1 & x_{31} & x_{32} \\
1 & x_{41} & x_{42}
\end{pmatrix}
\begin{pmatrix}
\beta_0 \\
\beta_1 \\
\beta_2
\end{pmatrix} +
\begin{pmatrix}
\varepsilon_1 \\
\varepsilon_2 \\
\varepsilon_3 \\
\varepsilon_4
\end{pmatrix}.
\]
\section{Parameter Estimation} \section{Parameter Estimation}
\subsection{Simple Linear Regression} \subsection{Simple Linear Regression}
@ -87,26 +113,26 @@ We want to minimize the distance between $\X\beta$ and $\Y$:
\Rightarrow& \forall i: \\ \Rightarrow& \forall i: \\
& \X_i \Y = \X_i X\hat{\beta} \qquad \text{where $\hat{\beta}$ is the estimator of $\beta$} \\ & \X_i \Y = \X_i X\hat{\beta} \qquad \text{where $\hat{\beta}$ is the estimator of $\beta$} \\
\Rightarrow& \X^\T \Y = \X^\T \X \hat{\beta} \\ \Rightarrow& \X^\T \Y = \X^\T \X \hat{\beta} \\
\Rightarrow& {\color{red}(\X^T \X)^{-1}} \X^\T \Y = {\color{red}(\X^T \X)^{-1}} (\X^T\X) \hat{\beta} \\ \Rightarrow& {\color{gray}(\X^\T \X)^{-1}} \X^\T \Y = {\color{gray}(\X^\T \X)^{-1}} (\X^\T\X) \hat{\beta} \\
\Rightarrow& \hat{\beta} = (X^\T\X)^{-1} \X^\T \Y \Rightarrow& \hat{\beta} = (X^\T\X)^{-1} \X^\T \Y
\end{align*} \end{align*}
This formula comes from the orthogonal projection of $\Y$ on the subspace define by the explanatory variables $\X$
This formula comes from the orthogonal projection of $\Y$ on the subspace define by the explicative variables $\X$
$\X \hat{\beta}$ is the closest point to $\Y$ in the subspace generated by $\X$. $\X \hat{\beta}$ is the closest point to $\Y$ in the subspace generated by $\X$.
If $H$ is the projection matrix of the subspace generated by $\X$, $X\Y$ is the projection on $\Y$ on this subspace, that corresponds to $\X\hat{\beta}$. If $H$ is the projection matrix of the subspace generated by $\X$, $X\Y$ is the projection on $\Y$ on this subspace, that corresponds to $\X\hat{\beta}$.
\section{Coefficient of Determination: \texorpdfstring{$R^2$}{R\textsuperscript{2}}}
\section{Coefficient of Determination: $R^2$}
\begin{definition}[$R^2$] \begin{definition}[$R^2$]
\[ \[
0 \leq R^2 = \frac{\norm{\X\hat{\beta} - \bar{\Y}\One}^2}{\norm{\Y - \bar{\Y}\One}^2} = 1 - \frac{\norm{\Y - \X\hat{\beta}}^2}{\norm{\Y - \bar{\Y}\One}^2} \leq 1 0 \leq R^2 = \frac{\norm{\X\hat{\beta} - \bar{\Y}\One}^2}{\norm{\Y - \bar{\Y}\One}^2} = 1 - \frac{\norm{\Y - \X\hat{\beta}}^2}{\norm{\Y - \bar{\Y}\One}^2} \leq 1
\] proportion of variation of $\Y$ explicated by the model. \] proportion of variation of $\Y$ explained by the model.
\end{definition} \end{definition}
\begin{figure}
\centering
\includestandalone{figures/schemes/orthogonal_projection}
\caption{Orthogonal projection of $\Y$ on plan generated by the base described by $\X$. $\color{blue}a$ corresponds to $\norm{\X\hat{\beta} - \bar{\Y}}^2$ and $\color{blue}b$ corresponds to $\norm{\Y - \hat{\beta}\X}^2$}
\label{fig:scheme-orthogonal-projection}
\end{figure}

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@ -15,15 +15,28 @@ Let $u = \begin{pmatrix}
v_n v_n
\end{pmatrix}$ \end{pmatrix}$
\begin{align*} \begin{definition}[Scalar Product (Dot Product)]
\langle u, v\rangle & = \left(u_1, \ldots, u_v\right) \begin{pmatrix} \begin{align*}
v_1 \\ \scalar{u, v} & = \begin{pmatrix}
\vdots \\ u_1, \ldots, u_v
v_n \end{pmatrix}
\end{pmatrix} \\ \begin{pmatrix}
& = u_1 v_1 + u_2 v_2 + \ldots + u_n v_n v_1 \\
\end{align*} \vdots \\
v_n
\end{pmatrix} \\
& = u_1 v_1 + u_2 v_2 + \ldots + u_n v_n
\end{align*}
We may use $\scalar{u, v}$ or $u \cdot v$ notations.
\end{definition}
\paragraph{Dot product properties}
\begin{itemize}
\item $\scalar{u, v} = \scalar{v, u}$
\item $\scalar{(u+v), w} = \scalar{u, w} + \scalar{v, w}$
\item $\scalar{u, v}$
\item $\scalar{\vec{u}, \vec{v}} = \norm{\vec{u}} \times \norm{\vec{v}} \times \cos(\widehat{\vec{u}, \vec{v}})$
\end{itemize}
\begin{definition}[Norm] \begin{definition}[Norm]
Length of the vector. Length of the vector.
@ -41,9 +54,7 @@ Let $u = \begin{pmatrix}
\end{definition} \end{definition}
\begin{definition}[Orthogonality] \begin{definition}[Orthogonality]
\[
u \perp v \Leftrightarrow \scalar{u, v} = 0
\]
\end{definition} \end{definition}
\begin{remark} \begin{remark}
@ -55,18 +66,17 @@ Let $u = \begin{pmatrix}
\] \]
\end{remark} \end{remark}
Scalar product properties: \begin{figure}
\begin{itemize} \centering
\item $\scalar{u, v} = \scalar{v, u}$ \includestandalone{figures/schemes/vector_orthogonality}
\item $\scalar{(u+v), w} = \scalar{u, w} + \scalar{v, w}$ \caption{Illustration for the scalar product of two orthogonal vectors.}
\item $\scalar{u, v}$ \label{fig:scheme-orthogonal-scalar-product}
\item $\scalar{\vec{u}, \vec{v}} = \norm{\vec{u}} \times \norm{\vec{v}} \times \cos(\widehat{\vec{u}, \vec{v}})$ \end{figure}
\end{itemize}
\begin{align*} \begin{align*}
\scalar{v-u, v-u} & = \scalar{v, v} + \scalar{u, u} - 2 \scalar{u, v} \\ \scalar{v-u, v-u} & = \scalar{v, v} + \scalar{u, u} - 2 \scalar{u, v} \\
& = \norm{v}^2 + \norm{u}^2 \\ & = \norm{v}^2 + \norm{u}^2 \\
& = -2 \scalar{u, v} & = -2 \scalar{u, v}
\end{align*} \end{align*}
\begin{align*} \begin{align*}
@ -74,10 +84,14 @@ Scalar product properties:
\norm{u + v}^2 & = \norm{u}^2 + \norm{v}^2 + 2 \scalar{u,v} \norm{u + v}^2 & = \norm{u}^2 + \norm{v}^2 + 2 \scalar{u,v}
\end{align*} \end{align*}
\begin{proposition}[Scalar product of orthogonal vectors]
\[
u \perp v \Leftrightarrow \scalar{u, v} = 0
\]
\end{proposition}
If $u \perp v$, then $\scalar{u, v} = 0$
\begin{proof}[Indeed] \begin{proof}[Indeed]
$\norm{u-v}^2 = \norm{u+v}^2$, $\norm{u-v}^2 = \norm{u+v}^2$, as illustrated in \autoref{fig:scheme-orthogonal-scalar-product}.
\begin{align*} \begin{align*}
\Leftrightarrow & -2 \scalar{u, v} = 2 \scalar{u, v} \\ \Leftrightarrow & -2 \scalar{u, v} = 2 \scalar{u, v} \\
\Leftrightarrow & 4 \scalar{u, v} = 0 \\ \Leftrightarrow & 4 \scalar{u, v} = 0 \\
@ -138,17 +152,17 @@ The scalar product between $z$ and (?) is zero.
Then, Then,
\begin{align*} \begin{align*}
Ax & = \begin{pmatrix} Ax & = \begin{pmatrix}
a & b \\ a & b \\
c & d c & d
\end{pmatrix} \end{pmatrix}
\begin{pmatrix} \begin{pmatrix}
x_1 \\ x_1 \\
x_2 x_2
\end{pmatrix} \\ \end{pmatrix} \\
& = \begin{pmatrix} & = \begin{pmatrix}
a x_1 + b_x2 \\ a x_1 + b x_2 \\
c x_1 + d x_2 c x_1 + d x_2
\end{pmatrix} \end{pmatrix}
\end{align*} \end{align*}
Similarly, Similarly,
@ -164,9 +178,9 @@ The scalar product between $z$ and (?) is zero.
x_3 \\ x_3 \\
x_4 x_4
\end{pmatrix} \end{pmatrix}
& = \begin{pmatrix} & = \begin{pmatrix}
a x_1 + b x_2 + c x_3 \ldots a x_1 + b x_2 + c x_3 \ldots
\end{pmatrix} \end{pmatrix}
\end{align*} \end{align*}
\end{example} \end{example}
@ -182,31 +196,8 @@ The number of columns has to be the same as the dimension of the vector to which
\end{pmatrix}$ \end{pmatrix}$
\end{definition} \end{definition}
\begin{example} \begin{figure}
\begin{align*} \centering
Y & = X \beta + \varepsilon \\ \includestandalone{figures/schemes/coordinates_systems}
\begin{pmatrix} \caption{Coordinate systems}
y_1 \\ \end{figure}
y_2 \\
y_3 \\
y_4
\end{pmatrix}
& = \begin{pmatrix}
1 & x_{11} & x_{12} \\
1 & x_{21} & x_{22} \\
1 & x_{31} & x_{32} \\
1 & x_{41} & x_{42}
\end{pmatrix}
\begin{pmatrix}
\beta_0 \\
\beta_1 \\
\beta_2
\end{pmatrix} +
\begin{pmatrix}
\varepsilon_1 \\
\varepsilon_2 \\
\varepsilon_3 \\
\varepsilon_4
\end{pmatrix}
\end{align*}
\end{example}

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@ -0,0 +1,16 @@
\documentclass[margin=0.5cm]{standalone}
\usepackage{tikz}
\usepackage{tkz-euclide}
\begin{document}
\usetikzlibrary{3d}
\begin{tikzpicture}
\tkzDefPoint(-2,-2){A}
\tkzDefPoint(10:3){B}
\tkzDefShiftPointCoord[B](1:5){C}
\tkzDefShiftPointCoord[A](1:5){D}
\tkzDrawPolygon(A,...,D)
\tkzDrawPoints(A,...,D)
\node at (A) {A};
\end{tikzpicture}
\end{document}

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@ -1,12 +1,23 @@
\documentclass[tikz]{standalone} \documentclass[tikz]{standalone}
\usepackage{tikz} \usepackage{tikz}
\usepackage{tkz-euclide}
\begin{document} \begin{document}
\begin{tikzpicture} \usetikzlibrary{3d}
\tkzInit[xmax=5,ymax=5,xmin=-5,ymin=-5] % 1D axis
\tkzGrid \begin{tikzpicture}[->]
\tkzAxeXY \begin{scope}[xshift=0]
\draw[thick, latex-latex] (-1,4) -- (4,-6) node[anchor=south west] {$a$}; \draw (0, 0, 0) -- (xyz cylindrical cs:radius=1) node[right] {$x$};
\end{scope}
% 2D coordinate system
\begin{scope}[xshift=50]
\draw (0, 0, 0) -- (xyz cylindrical cs:radius=1) node[right] {$x$};
\draw (0, 0, 0) -- (xyz cylindrical cs:radius=1,angle=90) node[above] {$y$};
\end{scope}
% 3D coordinate systems
\begin{scope}[xshift=100]
\draw (0, 0, 0) -- (xyz cylindrical cs:radius=1) node[right] {$x$};
\draw (0, 0, 0) -- (xyz cylindrical cs:radius=1,angle=90) node[above] {$y$};
\draw (0, 0, 0) -- (xyz cylindrical cs:z=1) node[below left] {$z$};
\end{scope}
\end{tikzpicture} \end{tikzpicture}
\end{document} \end{document}

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@ -0,0 +1,32 @@
% ref. https://tex.stackexchange.com/a/523362/235607
\documentclass[tikz]{standalone}
\usepackage{tikz-3dplot}
\usepackage{tkz-euclide}
\usepackage{mathtools}
\begin{document}
\tdplotsetmaincoords{50}{0}
\begin{tikzpicture}[tdplot_main_coords,bullet/.style={circle,inner
sep=1pt,fill=black,fill opacity=1}]
\begin{scope}[canvas is xy plane at z=0]
\tkzDefPoints{-2/-1/A,3/-1/B,4/2/C}
\tkzDefParallelogram(A,B,C)
\tkzGetPoint{D}
\tkzDrawPolygon[fill=gray!25!white](A,B,C,D)
\draw[decorate,decoration={brace,
amplitude=8pt},xshift=0pt,very thin,gray] (2,0) -- ++(-1,-0.5) node [black,midway,xshift=0.5em,yshift=-1em] {\color{blue}$a$};
\end{scope}
\begin{scope}[canvas is xz plane at y=0]
\draw[thick,fill=white,fill opacity=0.7,nodes={opacity=1}]
(2,0) node[bullet,label=below right:{$\mathbf{X}$}] {}
-- (0,0) node[bullet] {}
-- (0,3) node[bullet,label=above:{$\mathbf{Y}$}] {} -- cycle;
\draw (0.25,0) -- (0.25,0.25) -- (0,0.25);
\draw[decorate,decoration={brace,
amplitude=8pt},xshift=0pt,very thin,gray] (0,0) -- (0,3) node [black,midway,xshift=-1.25em,yshift=0em] {\color{blue}$b$};
\end{scope}
\begin{scope}[canvas is xy plane at z=0]
\draw[->] (2,0) -- ++(-0.75,0.75) node [left] {$\mathbf{1}$};
\draw[->] (2,0) -- ++(-1,-0.5);
\end{scope}
\end{tikzpicture}
\end{document}

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@ -0,0 +1,27 @@
\documentclass[margin=0.5cm]{standalone}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usepackage{mathtools}
\begin{document}
\begin{tikzpicture}
\coordinate (A) at (0.5, 1) {};
\coordinate (B) at (-0.5, -1) {};
\coordinate (C) at (1.25, -0.70) {};
\coordinate (0) at (0, 0) {};
% left angle
\tkzMarkRightAngle[draw=black,size=0.1](A,0,C);
\draw[lightgray,very thin] (A) -- (C);
% Curly brace annotation for ||u-v||
\draw[decorate,decoration={brace,
amplitude=10pt},xshift=0pt,yshift=4pt,very thin] (A) -- (C) node [black,midway,xshift=27pt,yshift=0.5em] {$\lVert u-v \rVert$};
\draw[lightgray,very thin] (B) -- (C);
% axis lines
\draw[->] (0) -- (A) node[above] {$u$};
\draw[->] (0) -- (B) node[below] {$-u$};
\draw[->] (0) -- (C) node[right] {$v$};
\end{tikzpicture}
\end{document}

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main.pdf

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@ -1,3 +1,6 @@
\usepackage{pgffor} \usepackage{pgffor}
\usetikzlibrary{math} \usetikzlibrary{math}
\usepackage{standalone} \usepackage{standalone}
\usepackage{tikz-3dplot}
\usepackage{tkz-euclide}
\usepackage{mathtools}